Correcting a Laplace Transform Problem

In summary: Yes, you are correct. After applying the limits, the first term becomes 0 and the second term becomes (s/(s^2 + a^2)). This is the correct answer.
  • #1
jayanthd
16
0

Homework Statement



Why I am getting wrong answer related to this Laplace Transforms problem?

According to the book "Higher Engineering Mathematics 6th edition by John O Bird" page no. 583 one should get

(e[itex]^{-st}[/itex]/(s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)

Homework Equations



∫e[itex]^{-st}[/itex]cos at dt

The Attempt at a Solution

u = e[itex]^{-st}[/itex]

du = -se[itex]^{-st}[/itex] dt

Let dv = cos at dt

v = (sin at)/ a

Integrating by parts

∫e[itex]^{-st}[/itex]cos at dt =

(e[itex]^{-st}[/itex] sin at / a) + (s/a)∫e[itex]^{-st}[/itex]sin at dt

= (e[itex]^{-st}[/itex] sin at / a) + (s/a)[(-e[itex]^{-st}[/itex] cos at / a) - (s/a)∫e[itex]^{-st}[/itex]cos at dt]

= (e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex] )(e[itex]^{-st}[/itex] cos at) - s[itex]^{2}[/itex]/a[itex]^{2}[/itex])∫e[itex]^{-st}[/itex]cos at dt]

Rearranging

(1 + (s[itex]^{2}[/itex]/a[itex]^{2}[/itex]))∫e[itex]^{-st}[/itex]cos at dt =

(e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex])(e[itex]^{-st}[/itex] cos at)

= (e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at)

∫e[itex]^{-st}[/itex]cos at dt =

(a[itex]^{2}[/itex]/(a[itex]^{2}[/itex] + s[itex]^{2}[/itex]))((e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at))

= ((e[itex]^{-st}[/itex]/ (s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)
 
Last edited:
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  • #2
jayanthd said:

Homework Statement



Why I am getting wrong answer related to this Laplace Transforms problem?

According to the book "Higher Engineering Mathematics 6th edition by John O Bird" page no. 583 one should get

(e[itex]^{-st}[/itex]/(s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)


Homework Equations



∫e[itex]^{-st}[/itex]cos at dt

The Attempt at a Solution




u = e[itex]^{-st}[/itex]

du = -se[itex]^{-st}[/itex] dt

Let dv = cos at dt

v = (sin at)/ a

Integrating by parts

∫e[itex]^{-st}[/itex]cos at dt =

(e[itex]^{-st}[/itex] sin at / a) + (s/a)∫e[itex]^{-st}[/itex]sin at dt

= (e[itex]^{-st}[/itex] sin at / a) + (s/a)[(-e[itex]^{-st}[/itex] cos at / a) - (s/a)∫e[itex]^{-st}[/itex]cos at dt]

= (e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex] )(e[itex]^{-st}[/itex] cos at) - s[itex]^{2}[/itex]/a[itex]^{2}[/itex])∫e[itex]^{-st}[/itex]cos at dt]

Rearranging

(1 + (s[itex]^{2}[/itex]/a[itex]^{2}[/itex]))∫e[itex]^{-st}[/itex]cos at dt =

(e[itex]^{-st}[/itex] sin at / a) - (s/a[itex]^{2}[/itex])(e[itex]^{-st}[/itex] cos at)

= (e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at)

∫e[itex]^{-st}[/itex]cos at dt =

(a[itex]^{2}[/itex]/(a[itex]^{2}[/itex] + s[itex]^{2}[/itex]))((e[itex]^{-st}[/itex]/a[itex]^{2}[/itex])(a sin at - s cos at))

= ((e[itex]^{-st}[/itex]/ (s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))(a sin at - s cos at)

There is no way you could get the answer that you claim the book obtains: the Laplace transform of a function f(t) will not have a "t" in it, since t has be "integrated out". I hope you realize that you need to take a limit!
 
  • #3
The integration limits on the LT integration are from 0 to ∞. Try these limits and see what you get. It's a definite integral, not an indefinite integral.
 
  • #4
Applying limits on the last step was not a problem. I was getting wrong answer that is there was a mistake in sign at one place where Integration by parts is done second time. After fixing it I got the right answer which I have modified in post 1. After applying the limits I got the right answer.

e[itex]^{-s\infty}[/itex] becomes 0

and

e[itex]^{-s * 0}[/itex] becomes 1

sin 0 = 0

cos 0 = 1

First term of the equation after applying infinity becomes 0. Remaining is applying 0 as limit but a - sign appears before the equation.

- [ e[itex]^{-s * 0}[/itex] / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex])(a sin a (0) - s cos a (0))

= - (1 / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex])) ( - s)

= (s / (s[itex]^{2}[/itex] + a[itex]^{2}[/itex]))

Am I right?
 
Last edited:

Related to Correcting a Laplace Transform Problem

1. What is a Laplace Transform problem?

A Laplace Transform problem is a mathematical problem that involves transforming a function from the time domain to the frequency domain using the Laplace Transform. This allows for the analysis of the behavior of a system in the frequency domain, which can provide insights that are not easily obtained in the time domain.

2. Why is the Laplace Transform used?

The Laplace Transform is used because it simplifies the analysis of complex systems. It allows for the conversion of differential equations, which are difficult to solve, into algebraic equations that are easier to manipulate and solve. This makes it a powerful tool in many areas of science and engineering.

3. How is the Laplace Transform applied?

The Laplace Transform is applied by taking the integral of a function multiplied by an exponential term. This integral can then be solved using known tables of Laplace Transform pairs or by using algebraic manipulation techniques. The resulting transformed function can then be analyzed in the frequency domain.

4. What are some common applications of the Laplace Transform?

The Laplace Transform has a wide range of applications in fields such as control theory, signal processing, electrical engineering, and physics. It is commonly used to analyze the stability and response of control systems, to solve differential equations in electrical circuits, and to study the behavior of mechanical systems.

5. Are there any limitations to the Laplace Transform?

While the Laplace Transform is a powerful tool, it does have some limitations. It can only be applied to functions with a finite number of discontinuities, and it is not suitable for functions that grow exponentially over time. In addition, the inverse Laplace Transform, which is used to convert a function back to the time domain, can be challenging to compute for some functions.

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